\[(a)\;\frac{1}{\alpha}+\frac{1}{\beta}=2\quad (b)\;\frac{1}{\alpha}+\frac{1}{\beta}=4\quad (c)\;\frac{1}{\alpha}+\frac{1}{\beta}=\frac{1}{2} \quad (d)\;\frac{1}{\alpha}+\frac{1}{\beta}=\frac{1}{4}\]

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Since $S_1+S_2=4 km$

and $t_1+t_2=4 min$

$t_1=\large\frac{v_{\Large max}}{\alpha}$

$t_2=\large\frac{v_{\Large max}}{\beta}$

$t_1+t_2=v_{max}\bigg(\large\frac{1}{\alpha}+\frac{1}{\beta}\bigg)$

$S_1=\large\frac{1}{2} v_{max} t_1$

$S_2=\large\frac{1}{2} v_{max} t_2$

$=>v_{\large max}=2$

Therefore $\large\frac{1}{\alpha}+\frac{1}{\beta}$$=2$

Hence a is the correct answer.

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